Module 2

Programming 8086

2.1 Motivation

Motivation of this course is to provide the basic knowledge of addressing modes

of 8086 microprocessor, instruction set of 8086, Assembly language programming ,

mixed language programming using stacks, strings, procedures , macros, timers,

counters & delay.

2.2 Syllabus

Module Contents Duration Self study

2.1 Addressing modes 2 Hours 2 Hours

2.2 Instruction Set 3 Hours 4Hours

2.3 Assembly Language Programming, 1Hours 2 Hours

2.4 Mixed Language Programming 1 Hours 2 Hours

2.5 Programs based on Stacks, Strings, Procedures, Macros, Timers, Counters & delay

3 Hours 5 Hours

2.3: Weightage:

2.4 Learning Objective / Learning Outcome

2.4.1 Learning Objective

Students will try analyzing processor performance improvement using instruction level

parallelism.

2.4.2 Learning Outcome

Students will be able to write assembly language program for given task for 8086

microprocessor.

2.5 Prerequisite:

2.6: Key Definition:

2.7 Introduction

2.7.1 Addressing modes Of 8086 Microprocessor

The 8086 microprocessor accesses the data in different ways such as from different registers, from memory locations or from I/O ports are called its addressing modes. These addressing modes are categorized according to the accessing method. These are as follows.

1. Register Addressing Modes (Accessing data from registers) 2. Immediate Addressing Modes (Accessing immediate data and storing in the

register as an operand ) 3. Memory Addressing Modes (Accessing data from memory) 4. Direct Addressing Modes (Accessing direct data from I/O port) 5. Again some instruction are classified according to their behavior or condition,

these are as follows. 6. Relative addressing modes (Related with some condition) 7. Implied or Implicit addressing mode (No operands)

1. Register Addressing Modes

In register addressing mode, most 8086 instructions can operate the general purpose register to set as an operand to the instruction. This means a register is a source of an operand as well as the register is only the destination of an operand for an instruction.

MOV destination, source; This instruction copies the data from the source location to the destination location. The 8-bit or 16 bit registers are certainly valid operands for this instruction. But both operands should be in the same size. Now let's see some 8086 MOV instructions: MOV AX, BX; Copies the 16 bit value from BX into AX, Here the contents of AX is overlapping, but the contents of BX are not changed. (Both registers are in same size)

Example:

MOV DL, AL; Copies the value from AL into DL MOV SI, DX; Copies the value from DX into SI MOV SP, BP; Copies the value from BP into SP MOV CH, CL; Copies the value from CL into CH MOV AX, AX; yes, this is legal!

2. Immediate Addressing Modes

In immediate addressing mode, the hexadecimal number either 8-bit or 16-bit to be loaded immediately into the memory locations or consecutive memory locations respectively. i.e to load 8-bit immediate number into an 8-bit memory location or to load a 16-bit immediate number into two consecutive memory locations. Some examples of that instructions.

Example:

MOV CX, 437BH; copies 16-bit hexadecimal number 437BH in the 16-bit CX register MOV CL, 48H; Load the 8-bit immediate number 48H into the 8-bit CL register

3. Memory Addressing Mode:

The memory addressing modes are used to specify the location of an operand as memory. To access data in memory, the 8086 should be produce a 20-bit physical address. It does this by adding a 16-bit value called the effective address (EA). The effective address represents the displacement or offset of the desired operand from the segment base. The data segment is most often used as a segment base. MOV AL, [BX]; ADD the contents of DS (One of four segment bases) with BX and result will shift in AL register. MOV AL, [BP]; same as above The square bracket around the registers i.e. BX, BP are shorthand for “the contents of that registers at a displacement from the segment base of ."

Fig 2.1

4. Direct Addressing modes

This addressing mode is called direct because the displacement or offset of the operand from the segment base is specified directly in the instruction. MOV CL, [437AH]; copy the contents of the memory location , at a displacement of 437AH from the data segment base , into the CL register. MOV BX [437AH]; copies a word from memory into the BX register. Note: each memory address represents a byte of storage, the word must come from two memory locations. The byte at a displacement of 437AH from the data segment base will be copied into BL. The contents of the next higher address, displacement 437BH will be copied into the BH register. The 8086 will automatically access the required number of bytes in memory for a given instruction.

5. Relative addressing modes

This addressing mode is a relation base. The data is stored either in base pointer (BP) or in BX. Example: MOV AX, [BP +1] JMP [BX + 1] JNC START : if CF=0 or not carry then start the PC

6. Implicit or Implied addressing modes

In this addressing mode no operands are used to execute the instruction. Example: NOP : No operation CLC : Clear carry flag to 0 START : Start execution

2.7.2 Instruction Set of 8086

Instruction set of 8086 microprocessor is briefed in below table

AAA AAD AAM AAS ADC ADD AND CALL CBW CLC CLD CLI CMC CMP

CMPSB CMPSW CWD DAA DAS DEC DIV HLT IDIV IMUL IN INC INT INTO IRET JA

JAE JB JBE JC JCXZ JE JG JGE JL JLE JMP JNA JNAE JNB

JNBE JNC JNE JNG JNGE JNL JNLE JNO JNP JNS JNZ JO JP JPE

JPO JS JZ LAHF LDS LEA LES LODSB LODSW LOOP LOOPE LOOPNE LOOPNZ LOOPZ

MOV MOVSB MOVSW MUL NEG NOP NOT OR OUT POP POPA POPF PUSH PUSHA PUSHF RCL

RCR REP REPE REPNE REPNZ REPZ RET RETF ROL ROR SAHF SAL SAR SBB

SCASB SCASW SHL SHR STC STD STI STOSB STOSW SUB TEST XCHG XLATB XOR

The 8086 instructions are categorized into the following main types. i. Data Copy / Transfer Instructions ii. Arithmetic and Logical Instructions iii. Branch Instructions iv. Loop Instructions v. Machine Control Instructions vi. Flag Manipulation Instructions vii. Shift and Rotate Instructions viii. String Instructions

Data Copy / Transfer Instructions : MOV :

This instruction copies a word or a byte of data from some source to a destination. The destination can be a register or a memory location. The source can be a register, a memory location, or an immediate number. MOV AX,BX MOV AX,5000H MOV AX,[SI] MOV AX,[2000H] MOV AX,50H[BX]

MOV [734AH],BX MOV DS,CX MOV CL,[357AH] Direct loading of the segment registers with immediate data is not permitted.

PUSH : Push to Stack

This instruction pushes the contents of the specified register/memory location on to the stack. The stack pointer is decremented by 2, after each execution of the instruction. E.g. PUSH AX • PUSH DS • PUSH [5000H]

Fig 2.2

POP : Pop from Sack

This instruction when executed, loads the specified register/memory location with the contents of the memory location of which the address is formed using the current stack segment and stack pointer. The stack pointer is incremented by 2 Eg. POP AX POP DS POP [5000H]

Fig 2.3

XCHG : Exchange byte or word

This instruction exchange the contents of the specified source and destination operands Eg. XCHG [5000H], AX XCHG BX, AX

XLAT :

Translate byte using look-up table Eg. LEA BX, TABLE1 MOV AL, 04H XLAT Simple input and output port transfer Instructions:

IN:

Copy a byte or word from specified port to accumulator. Eg. IN AL,03H IN AX,DX

OUT:

Copy a byte or word from accumulator specified port. Eg. OUT 03H, AL OUT DX, AX

LEA :

Load effective address of operand in specified register.[reg] offset portion of address in DS Eg. LEA reg, offset

LDS:

Load DS register and other specified register from memory. [reg] [mem] [DS] [mem + 2] Eg. LDS reg, mem

LES:

Load ES register and other specified register from memory. [reg] [mem] [ES] [mem + 2] Eg. LES reg, mem

2.7.3 Assembly Language Programming

Assembly Language:

Each personal computer has a microprocessor that manages the computer's

arithmetical, logical, and control activities.

Each family of processors has its own set of instructions for handling various

operations such as getting input from keyboard, displaying information on screen and

performing various other jobs. These set of instructions are called 'machine language

instructions'.

A processor understands only machine language instructions, which are strings

of 1's and 0's. However, machine language is too obscure and complex for using in

software development. So, the low-level assembly language is designed for a specific

family of processors that represents various instructions in symbolic code and a more

understandable form.

Advantages of Assembly Language

Having an understanding of assembly language makes one aware of −

How programs interface with OS, processor, and BIOS;

How data is represented in memory and other external devices;

How the processor accesses and executes instruction;

How instructions access and process data;

How a program accesses external devices.

Other advantages of using assembly language are −

It requires less memory and execution time;

It allows hardware-specific complex jobs in an easier way;

It is suitable for time-critical jobs;

It is most suitable for writing interrupt service routines and other memory

resident programs.

Basic Features of PC Hardware

The main internal hardware of a PC consists of processor, memory, and

registers. Registers are processor components that hold data and address. To execute a

program, the system copies it from the external device into the internal memory. The

processor executes the program instructions.

The fundamental unit of computer storage is a bit; it could be ON (1) or OFF (0).

A group of nine related bits makes a byte, out of which eight bits are used for data and

the last one is used for parity. According to the rule of parity, the number of bits that

are ON (1) in each byte should always be odd.

So, the parity bit is used to make the number of bits in a byte odd. If the parity is

even, the system assumes that there had been a parity error (though rare), which might

have been caused due to hardware fault or electrical disturbance.

The processor supports the following data sizes −

Word: a 2-byte data item

Double word: a 4-byte (32 bit) data item

Quad word: an 8-byte (64 bit) data item

Paragraph: a 16-byte (128 bit) area

Kilobyte: 1024 bytes

Megabyte: 1,048,576 bytes

Addressing Data in Memory

The process through which the processor controls the execution of instructions is

referred as the fetch-decode-execute cycle or the execution cycle. It consists of three

continuous steps −

Fetching the instruction from memory

Decoding or identifying the instruction

Executing the instruction

The processor may access one or more bytes of memory at a time. Let us consider a

hexadecimal number 0725H. This number will require two bytes of memory. The high-

order byte or most significant byte is 07 and the low-order byte is 25.

The processor stores data in reverse-byte sequence, i.e., a low-order byte is stored in a

low memory address and a high-order byte in high memory address. So, if the

processor brings the value 0725H from register to memory, it will transfer 25 first to

the lower memory address and 07 to the next memory address.

Fig 2.4 memory address

When the processor gets the numeric data from memory to register, it again reverses

the bytes. There are two kinds of memory addresses −

Absolute address - a direct reference of specific location.

Segment address (or offset) - starting address of a memory segment with the

offset value.

An assembly program can be divided into three sections −

The data section,

The bss section, and

The text section.

The data Section

The data section is used for declaring initialized data or constants. This data does not

change at runtime. You can declare various constant values, file names, or buffer size,

etc., in this section.

The syntax for declaring data section is −

section.data

The bss Section

The bss section is used for declaring variables. The syntax for declaring bss section is −

section.bss

The text section

The text section is used for keeping the actual code. This section must begin with the

declaration global _start, which tells the kernel where the program execution begins.

The syntax for declaring text section is −

section.text global _start _start:

Comments

Assembly language comment begins with a semicolon (;). It may contain any printable

character including blank. It can appear on a line by itself, like −

; This program displays a message on screen

or, on the same line along with an instruction, like −

add eax, ebx ; adds ebx to eax

Assembly Language Statements

Assembly language programs consist of three types of statements −

Executable instructions or instructions,

Assembler directives or pseudo-ops, and

Macros.

The executable instructions or simply instructions tell the processor what to do. Each

instruction consists of an operation code (opcode). Each executable instruction

generates one machine language instruction.

The assembler directives or pseudo-ops tell the assembler about the various aspects of

the assembly process. These are non-executable and do not generate machine language

instructions.

Macros are basically a text substitution mechanism.

Syntax of Assembly Language Statements

Assembly language statements are entered one statement per line. Each statement

follows the following format −

[label] mnemonic [operands] [;comment]

The fields in the square brackets are optional. A basic instruction has two parts, the

first one is the name of the instruction (or the mnemonic), which is to be executed, and

the second are the operands or the parameters of the command.

Following are some examples of typical assembly language statements −

INC COUNT ; Increment the memory variable COUNT MOV TOTAL, 48 ; Transfer the value 48 in the ; memory variable TOTAL ADD AH, BH ; Add the content of the ; BH register into the AH register AND MASK1, 128 ; Perform AND operation on the ; variable MASK1 and 128 ADD MARKS, 10 ; Add 10 to the variable MARKS MOV AL, 10 ; Transfer the value 10 to the AL register

The Hello World Program in Assembly

The following assembly language code displays the string 'Hello World' on the screen

−

section .text

global _start ;must be declared for linker (ld)

_start: ;tells linker entry point

mov edx,len ;message length

mov ecx,msg ;message to write

mov ebx,1 ;file descriptor (stdout)

mov eax,4 ;system call number (sys_write)

int 0x80 ;call kernel

mov eax,1 ;system call number (sys_exit)

int 0x80 ;call kernel

section .data

msg db 'Hello, world!', 0xa ;string to be printed

len equ $ - msg ;length of the string

When the above code is compiled and executed, it produces the following result −

Hello, world!

Compiling and Linking an Assembly Program in NASM

Make sure you have set the path of nasm and ld binaries in your PATH environment

variable. Now, take the following steps for compiling and linking the above program −

Type the above code using a text editor and save it as hello.asm.

Make sure that you are in the same directory as where you saved hello.asm.

To assemble the program, type nasm -f elf hello.asm

If there is any error, you will be prompted about that at this stage. Otherwise, an

object file of your program named hello.o will be created.

To link the object file and create an executable file named hello, type ld -m

elf_i386 -s -o hello hello.o

Execute the program by typing ./hello

If you have done everything correctly, it will display 'Hello, world!' on the screen.

2.7.4 Mixed Language Programming:

1. There are times when programs need to call programs written in other languages referred as mixed language programming. For example, when a particular subprogram is available in a language other than language you are using, or when algorithms are described more naturally in a different language, you need to use more than one language.

2. Mixed-language programming always involves a call to a function, procedure, or subroutine. Mixed-language calls involve calling functions in separate modules. Instead of compiling all source programs with same compiler, different compilers or assemblers are used as per the language used in the programs.

3. Microsoft C supports this mixed language programming. So it can combine assembly code routines in C as a separate language.

4. C program calls assembly language routines that are separately assembled by-MASM (MASM Assembler). These assembled modules are linked with the compiled C modules to get executable file. Fig shows the compile, assemble and link processes using C compiler, MASM assembler, and TUNIC.

Fig 2.5

2.7.5 Programs based on Stacks, Strings, Procedures, Macros, Timers, Counters &

delay

1.Write and execute an Assembly Language Program (ALP) to 8086 Processor to Add,

Subtract and Multiply 16 bit unsigned numbers. Store the result in extra segment.

ALGORITHM:

1. Start the Data segment

2. Initialize the input1 with 16-bit number and input2 with 16-bit number

3. End the Data segment

4. Start the extra segment

5. End extra segment

6. Assume data, code, extra segments for assembler

7. Initialize Data segment & extra segment

8. Refresh the Contents of AX register with XOR operation

9. Move input1 to AX register

10. Move input2 to BX register

11. Perform addition/subtraction/division operation

12. Move the result to corresponding locations

13. End the Code segment & the program

DATA SEGMENT

Value1 DW1234H

Value2 DW0123H

DATA ENDS

ASSUME CS:CODE, DS:DATA

CODE SEGMENT

START: MOV AX, DATA

MOV DS,AX

MOV AX,0000H

MOV BX,0000H

MOV SI,OFFSET Value1

MOV DI,OFFSET Value2

MOV AX,[SI]

MOV BX,[DI]

ADD AX, BX

INT 3H

CODE ENDS

END START

2. Write and execute an Assembly Language Program (ALP) to 8086 Processor to sort the

given array in ascending & descending order.

ALGORITHM:

1. Start the Data segment

2. Initialize the input values by using Assembler directives

3. Ends the Data segment

4. Start the code segment

5. Initialize START Directive

6. Initialize Data segment

7. Clear the Contents of AX, & BX registers.

8. Move the array values into SI register.

9. Initialize the count register & Move the first content of SI register to AL register

10. Increment SI value by one.

11. Compare the AL content with content of SI register.

12. If result is Positive perform exchange operation between AL register & content of SI register.

13. Decrement SI by one& Move the content of AL to SI location.

14. Increment SI register & Decrement count register by one.

15. Perform conditional statement JNZ LOOP1

16. Decrement BX register by one

17. Perform conditional statement JNZ LOOP2

18. Terminate the program with Interrupt statement.

19. Ends the code segment.

DATA SEGMENT

Num1 DB 52H, 30H, 62H, 42H, 17H

DATA ENDS

ASSUME CS:CODE, DS:DATA

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV AX, 0000H

MOV BX, 0000H

BACK: MOV SI, OFFSET Num1

MOV CX, 0004H

BACK1: MOV AL, [SI]

INC SI

CMP AL, [SI]

JBE X

XCHG AL, [SI]

DEC SI

MOV [SI], AL

INC SI

X: DEC CX

JNZ BACK1

DEC BX

JNZ BACK

INT 3H

CODE ENDS

END START

DESCENDING ORDER

DATA SEGMENT

Num1 DB 52H, 30H, 62H, 42H, 17H

DATA ENDS

ASSUME CS: CODE, DS: DATA

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV AX, 0000H

MOV BX, 0000H

BACK: MOV SI, OFFSET Num1

MOV CX, 0004H

BACK1: MOV AL, [SI]

INC SI

CMP AL, [SI]

JNBE X

XCHG AL, [SI]

DEC SI

MOV [SI], AL

INC SI

X: DEC CX

JNZ BACK1

DEC BX

JNZ BACK

INT 3H

CODE ENDS

END START

3.To Write and execute an Assembly Language Program (ALP) to 8086 Processor to find the

length of a given string which terminates with a special character‘$’.

ALGORITHM:

1. Start the Data segment

2. Initialize the given string in data segment which is ended with special character.

3. Ends the Data segment

4. Start the code segment

5. Initialize START Directive

6. Initialize Data segment

7. Clear the Contents of AX register.

8. Initialize the CX registers as zero.

9. Assign the values to SI register using indexed addressing mode.

10. Move the content of SI to AL register

11. Compare AL with “$”

12. Perform Conditional Operation if JE them jump to Interrupt

13. Else Increment SI & Increment CX

14. Repeat the loop until Compare with $ symbol.

15. Divide the result by the content of BX register.

16. Terminate the program with Interrupt statement

17. Ends the Code segment.

18. End the program.

DATA SEGMENT

Str DB “INDIA$”

DATA ENDS

ASSUME CS: CODE, DS: DATA

CODE SEGMENT

START: MOV AX, DATA

MOV DS, AX

MOV AX, 0000H

MOV CX, 0000H

MOV SI, OFFSET Str

X: MOV AL, [SI]

CMP AL, ‘$’

JE Y

INC SI

INC CX

JMP X

Y: INT 3H

CODE ENDS

END START

2.8 References :

1. Microprocessors and Interfacing ,Douglas V Hall,T ata Mc Graw Hill

2. http://www.telnet7.net/articles/not_mine/8086_achitecture.htm

3. The 8086/8088 Family, John Uffenbuck, Pearson Media, LPE

2.9 Objective Questions

1. Which of the following is an illegal 8086 instruction?

(a)MOV 20h,BX (b) INC AL (c) AND BX,BX (d) ADD AX,30h

2. Which of the following is an illegal 8086 instruction

(a) MOV AX,[BX] (b) INC [BX] (c) ADD [AX], [BX] (d) ADD AX, [CX]

3..Branch prediction is used in the context of

(a) pipelining (b) program loops (c) cache memory (d) ALU operation

4. A conditional jump instruction

(a) always cause a transfer of control

(b) always involves the use of the status register

(c) always modifies the program counter

(d) always involves testing the Zero flag

5. The advantage of memory mapped I/O over I/O mapped I/O is,

(a) Many instructions supporting memory mapped I/O (b) Faster

(c) Require a bigger address decoder (d) All the above

6. BHE(bar) of 8086 microprocessor signal is used to interface the

(a) Even bank memory (b) Odd bank memory

(c) I/O (d) DMA

7. In 8086 microprocessor one of the following statements is not true.

(a) Coprocessor is interfaced in MAX mode

(b) Coprocessor is interfaced in MIN mode

(c) I/O can be interfaced in MAX / MIN mode

(d) Supports pipelining

8. What is meant by Maskable interrupts?

(a) An interrupt which can never be turned off.

(b) An interrupt that can be turned off by the programmer.

(c)both a& b

(d) none

9. In 8086 the overflow flag is set when

(a) The sum is more than 16 bits

(b) Signed numbers go out of their range after an arithmetic operation

(c) Carry and sign flags are set

(d) During subtraction

10. In 8086 microprocessor the following has the highest priority among all type interrupts.

(a) NMI (b) DIV 0 (c) TYPE 255 (d) OVER FLOW

Answer Key:-

1. A 2 C 3 A 4 A 5 A 6 B 7 B 8 B 9 B 10 A

2.10 Short Questions

1. What are the various interrupts in 8086?

2. What is meant by Maskable interrupts?

03.What is Non-Maskable interrupts?.

04.Which interrupts are generally used for critical events?

05.Give example for Non-Maskable interrupts?

06.What is the difference between microprocessor and microcontroller?

07. What is interrupt?

2.11 Subjective Questions (Based on University Patterns)

1. What is addressing mode? Explain different modes supported by 8086 with examples.

2. Explain memory organization concept of 8086.

3. Explain memory banking concept of 8086?

4. Write a procedure to calculate the factorial of the N using recursive procedure.

5. Differentiate between: procedure and macro.

6. WAP for 8086 in assembly language to check if string initialized in the data segment is

Palindrome or not.

7. Explain with example string instructions.

8. WAP to add 4 digit BCD number in AX to the similar in BX maintaining valid BCD result.

9. What do you mean by assembler directive? Explain any four in detail?

10. Explain the following instructions:

a. XLAT

b. RCL

c. LAHF

d. TEST

e. SAR

f. DAS

g. AAD

h. DAA

i. CLD

j. MOVSB

11. Explain interrupt structure of 8086.

12. What is interrupt? Explain different types of interrupts supported by 8086.

13. Write short notes on I/O mapped I/O and memory mapped I/O

14. Design 8086 based single board microprocessor system with following specification:

a. 8086 CPU working at 5 MHz

b. 64k of EPROM memory using 16K devices for operating software.

c. 256K of SRAM using 64k devices

d. Two input ports and two output ports [all 8-bit], interrupt driven.

Explain the design. Show memory and I/O mapped.

15. Design 8086 based single board microprocessor system with following specification:

A. 8086 CPU working at 5 MHz

B. 64k of EPROM memory using 16K devices

C. 128K of SRAM using 64k devices